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4-2.Quadratic Equations and Inequations
hard
The product of all the rational roots of the equation $\left(x^2-9 x+11\right)^2-(x-4)(x-5)=3$, is equal to :
A$14$
B$7$
C$28$
D$21$
(JEE MAIN-2025)
Solution
$\left(x^2-9 x+11\right)^2-\left(x^2-9 x+20\right)=3$
Let
$\Rightarrow x^2-9 x=t$
$\Rightarrow t^2+22 t+121-t-20-3=0$
$\Rightarrow t^2+21 t+98=0$
$\Rightarrow(t+14)(t+7)=0$
$\Rightarrow t=-7,-14$
So,
$x^2-9 x=-7,-14 $
$x^2-9 x+7=0 \text { or } x^2-9 x+14=0$
$x=\frac{9 \pm \sqrt{81-4(7)}}{2 \times 1} x=\frac{9 \pm \sqrt{81-4(14)}}{2}$
$=\frac{9 \pm \sqrt{53}}{2} =\frac{9 \pm 5}{2}$
Product of all rational roots $=7 \times 2=14$
Let
$\Rightarrow x^2-9 x=t$
$\Rightarrow t^2+22 t+121-t-20-3=0$
$\Rightarrow t^2+21 t+98=0$
$\Rightarrow(t+14)(t+7)=0$
$\Rightarrow t=-7,-14$
So,
$x^2-9 x=-7,-14 $
$x^2-9 x+7=0 \text { or } x^2-9 x+14=0$
$x=\frac{9 \pm \sqrt{81-4(7)}}{2 \times 1} x=\frac{9 \pm \sqrt{81-4(14)}}{2}$
$=\frac{9 \pm \sqrt{53}}{2} =\frac{9 \pm 5}{2}$
Product of all rational roots $=7 \times 2=14$
Standard 11
Mathematics
